Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<bitset>
#include<set>
#define N (200010)
#define P ()
#define M ()
#define inf (0x7f7f7f7f)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
namespace fastIO1{
	inline char read(){
		static const int IN_LEN=1000000;
		static char buf[IN_LEN],*s,*t;
		return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
	}
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=read();!isdigit(c);c=read()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
	inline char readc(char &c){
		for(c=read();!isalpha(c)&&!isdigit(c);c=read())
		if(c==-1)return 0;
	}
	const int OUT_LEN = 10000000;
	char obuf[OUT_LEN],*ooh=obuf;
	inline void print(char c) {
		if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
		*ooh++=c;
	}
	template<class T>
	inline void print(T x){
		static int buf[30],cnt;
		if(x==0)print('0');
		else{
			if(x<0)print('-'),x=-x;
			for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
			while(cnt)print((char)buf[cnt--]);
		}
	}
	inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
}
namespace fastIO2{
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=getchar();!isdigit(c);c=getchar()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=getchar())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
}
using namespace fastIO1;
int n,a[N],pos[N],T[N],f[N];
int main(){
	read(n); for(int i=1;i<=n;i++)read(a[i]),pos[a[i]]=i;
	for(int i=n-1;i;i--){
		if(T[i+1]==0){
			if(pos[i]>pos[i+1])T[i]=1,f[i]=i+1;
			else T[i]=0;
		}
		else{
			int s=(pos[f[i+1]]<pos[i])+(pos[i]<pos[i+1])+(pos[i+1]<pos[f[i+1]]);
			if(s==2)T[i]=T[i+1],f[i]=f[i+1];
			else T[i]=T[i+1]+1,f[i]=i+1;
		}
	}
	cout<<T[1]<<endl;
}